College Algebra Exam Review 127

# College Algebra Exam Review 127 - b.T A b and furthermore...

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2.7. QUOTIENT GROUPS AND HOMOMORPHISM THEOREMS 137 The inverse of T A; b is T A ± 1 ; ± A ± 1 b . N is a subgroup isomorphic to the additive group R n because T E; b T E; b 0 D T E; b C b 0 ; and N is normal. In fact, T A; b T E; c T ± 1 A; b D T E;A c : Let us examine the condition for two elements T A; b and T A 0 ; b 0 to be con- gruent modulo N . The condition is T ± 1 A 0 ; b 0 T A; b D T A 0 ± 1 ; ± A 0 ± 1 b 0 T A; b D T A 0 ± 1 A;A 0 ± 1 . b ± b 0 / 2 N: This is equivalent to A D A 0 . Thus the class of T A; b modulo N is ŒT A; b Ł D f T A; b 0 W b 0 2 R n g , and the cosets of N can be parameterized by A 2 GL .n/ . In fact, the map ŒT A; b Ł 7! A is a bijection between the set Aff .n/=N of cosets of Aff .n/ modulo N and GL .n/ . Let us write ' for the map ' W T A; b 7! A from Aff .n/ to GL .n/ , and Q ' for the map Q ' W ŒT A; b Ł 7! A from Aff .n/=N to GL .n/ . The map ' is a (surjective) homomorphism, because '.T A; b T A 0 ; b 0 / D '.T AA 0 ;A b 0 C b / D AA 0 D '.T A;
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Unformatted text preview: b /'.T A ; b /; and furthermore the kernel of ' is N . The deﬁnition of the product in Aff .n/=N is ŒT A; b ŁŒT A ; b Ł D ŒT A; b T A ; b Ł D ŒT AA ; b C A b Ł: It follows that Q '.ŒT A; b ŁŒT A ; b Ł/ D Q '.ŒT AA ; b C A b Ł/ D AA D Q '.ŒT A; b Ł/ Q '.ŒT A ; b Ł/; and, therefore, Q ' is an isomorphism of groups. We can summarize our ﬁndings in the diagram: Aff .n/ ' q q q q q GL .n/ ± q q q q q ± ± ± ± ± ± q q q q q q q ∼ = Q ' Aff .n/=N...
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